Решите систему уравнений: 2x-y=10; x^2-3xy-y^2=4.

\[- x^{2} - 10x + 96 = 0\]

\[x^{2} + 10x - 96 = 0\]

\[D = 10^{2} - 4 \cdot 1 \cdot ( - 96) =\]

\[= 100 + 84 = 484;\ \ \ \ \sqrt{D} = 22.\]

\[x_{1} = \frac{- 10 + 22}{2} = \frac{12}{2} = 6;\ \ \ \]

\[\text{\ \ }x_{2} = \frac{- 10 - 22}{2} = \frac{- 32}{2} = - 16\]

\[y_{1} = 2 \cdot 6 - 10 = 12 - 10 = 2;\]

\[y_{2} = 2 \cdot ( - 16) - 10 =\]

\[= - 32 - 10 = - 42.\]

\[Ответ:(6;2);\ \ ( - 16;\ - 42).\]